JEE Advanced 2022 — Paper 2 — PHY-II solution

JEE Advanced2022Paper 2PHY-II

In Circuit-1 and Circuit-2 shown in the figures, $R_{1} = 1 Ω$ , $R_{2} = 2 Ω$ and $R_{3} = 3 Ω$ .

$P_{1}$ and $P_{2}$ are the power dissipations in Circuit-1 and Circuit-2 when the switches $S_{1}$ and $S_{2}$ are in open conditions, respectively.

$Q_{1}$ and $Q_{2}$ are the power dissipations in Circuit-1 and Circuit-2 when the switches $S_{1}$ and $S_{2}$ are in closed conditions, respectively.

Which of the following statement(s) is(are) correct?

A
When a voltage source of 6 V is connected across A and B in both circuits, $P_{1} < P_{2}$
B
When a constant current source of 2 Amp is connected across A and B in both circuits, $P_{1} > P_{2}$
C
When a voltage source 6 V is connected across A and B in Circuit-1, $Q_{1} > P_{1}$
D
When a constant current source of 2 Amp is connected across A and B in both circuits, $Q_{2} < Q_{1}$

Solution

When $S_{1}$ and $S_{2}$ are open

$(R_{eq})_1 = 1+\dfrac{5\times\dfrac{1}{2}} {5+\dfrac{1}{2}} = 1+\dfrac{5}{11} = \dfrac{16}{11}$

$P_1 = \dfrac{V^2}{R_{eq}} = \dfrac{(6)^2}{16/11} = \dfrac{36\times11}{16} = 24.75\ \text{W}$

$(R_{e q})_{2} = \frac{6}{11} Ω$

$P_2 = \dfrac{V^2}{R_{eq}} = \dfrac{(6)^2}{6/11} = \dfrac{36\times11}{6} = 66\ \text{W}$

$P_{2} > P_{1}$

Option (A) is correct.

$\Rightarrow$ If a $2 A$ source is used in both the cases,

$P_1 = i^2(R_{eq})_1 = (2)^2\times\dfrac{16}{11} = \dfrac{64}{11} = 5.818\ \text{W}$

$P_2 = i^2(R_{eq})_2 = (2)^2\times\dfrac{6}{11} = \dfrac{24}{11} = 2.1818\ \text{W}$

$P_{1} > P_{2}$

Option (B) is correct.

For $Q_{1}$

$R_{e q} = \frac{5}{11} Ω$

$Q_1 = \dfrac{V^2}{R_{eq}} = \dfrac{(6)^2}{5/11} = \dfrac{36\times11}{5} = 79.2\ \text{W}$

$P_{1} = 24.75 W$

$Q_{1} > P_{1}$

Option (C) is correct.

For option (D)

$Q_1 = i^2R_{eq} = (2)^2\times\dfrac{5}{11} = \dfrac{20}{11} = 1.81\ \text{W}$

$Q_2 = i^2R_{eq} = (2)^2\times\dfrac{1}{2} = \dfrac{4}{2} = 2\ \text{W}$

$Q_{2} > Q_{1}$

Option (D) is incorrect.

$∴$ Correct options: (A), (B), (C).

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